Crack the Problem

(5) Evaluate

Submission: Closed

Cracked by:

1. **Tarun Arora** (The first to crack), NIT Surathkal. Tarun has cleared mains 2003, congrats!
2. Ravi Subramanian, IIT Bombay, B.Tech. 1'st Year.
3. Tushar Jain, Meerut Public School, Meerut U.P.
4. Sankalp Guha, Vivekananda Kendra Vidyalaya, Port Blair.
5. M. Chetan Reddy, New Generation Junior College, Class XII, Hyderabad.

(4)

A particle of mass m is placed on a frictionless horizontal surface. It is attached to the ends of three horizontal springs S1, S2 and S3 of equal stiffness k. The particle is in equilibrium with S2 and S3 at right angles and S1 making equal angles with S2 and S3. If the particle is slightly pushed against S1 and released find its period of oscillation.

Submission: Closed

Cracked by:

1. Mayank Gaur.
2. Tushar Jain, Meerut Public School, Meerut U.P.
3. Nishant Kumar, Class XII, Modern School, Nagpur.
4. Anupam Narayan, Class XII, Swami Vivekanand Junior College, Mumbai.
5. Umang Merwana, Satna (M.P.).

(3) No three diagonals of a convex decagon intersect at one point. The vertices of the decagon constitute the set A and the intersection points of the diagonals constitute the set B. 3 elements are picked at random from the set A U B to form a set C. What is the chance that C is a subset of B?

Submission: Closed

Cracked by:

Nachiket Vasant Vaidya, 12th, Sathaye College, Vile Parle, Mumbai.

Solution Provided by Nachiket

As no three diagonals intersect each other, If we consider any for vertices of the polygon there will be only one intersection of the vertices the other drawn diagonals will not correspond to intersections.

Hence the number of intersection points are ⁿC₄ i.e. ¹⁰C₄ = 210.
Hence, n(A)=10 and n(B)=210 and n(A U B)=220
Hence the required Probability is
P= ²¹⁰C₃/²²⁰C₃= 7*19*52/73*109 =6916/7957=0.869

(2) If a + b + g =90°, prove that sin²a + sin²b + sin²g + 2 sina sinb sing =1.

Submission: Closed

Cracked by:

1. **Dhanush Kandathil** (first to crack), Trivandrum.
2. Ramesh Nidadavolu, Hyderabad.
3. Sachin Aggarwal, Delhi.

Here is the solution provided by Dhanush Kandathil:

a + b + g= 90.
This means sing is the same as cos(a + b).

Substituting in Sigma(sin²a) + 2PI(sin a) = sin²a + sin²b + cos²(a + b) + 2sinasinbcos(a+b)
but cos(a+b)=cosacosb -sinasinb
cos²(a+b)= cos²acos²b +sin²asin²b - 2sinasinbcosacosb

Also We have
2sinasinbcos(a+b)=2sinasinbcosacosb-2sin²asin²b

On adding the 2sinasinbcosacosb terms cancel out.
and expression reduces to sin²a+sin²b+ (cosacosb)²-(sinasinb)²

But(cosacosb)²-(sinasinb)² = cos(a+b)cos(a-b)=cos²b-sin²a +cos²b = 1, which is the required answer.

(1) \(\int cos^8 x dx\)

Cracked by:

Rahul Deshpande from Bhopal.

Submission: Closed

Here is the solution provided by him:

f(x) = cos⁸(x)

Using the method of integration by DeMoivre's theorem.

from DeMoivre's theorem,
\(z+{1\over z}=2\cos x\)
also, \(z^n +{1\over {z^n }}=2\cos nx\) --> (1), where z is a complex number

therefore,
\((2\cos x)^8 =\left({z+{1\over z}}\right)^8\)
\(256\cos ^8 x =z^8 +{1\over {z^8 }}+8\left({z^6 +{1\over {z^6 }}}\right)+28\left({z^4 + {1\over {z^4 }}}\right)+56\left({z^2 +{1\over {z^2 }}}\right)+70\)

therefore, final answer after using (1)\(\int {\cos ^8 xdx={1\over {128}}\left[{{1\over 8}\sin 8x+{4\over 3}\sin 6x+7\sin 4x+28\sin 2x+35x}\right]}+C\)